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what is 4" dai length

what is the 4" dai length

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Suggested Questions And Answer :

what is 4" dai length

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What is the diameter of a spiral coil of .65265 inch diameter pipe 100 feet long?

The equation of a spiral in polar coordinates has the general form r=A+Bø, where A is the starting radius of the spiral and B is a factor governing the growth of the spiral outwards. For example, if B=0, there is no outward growth and we just have a circle of radius A. A horizontal line length A represents the initial r, and the angle ø is the angle between r and this horizontal line. So r increases in length as ø increases (this angle is measured in radians where 2(pi) radians = 360 degrees, so 1 radian is 180/(pi)=57.3 degrees approximately.) If B=1/2 and A=5", for example, the minimum radius would be 5" when ø=0. When ø=2(pi) (360 degrees), r=5+(pi), or about 8.14". This angle would bring r back to the horizontal position, but it would be 8.14" instead of the initial 5". At ø=720 degrees, the horizontal line would increase by a further 3.14". Everywhere on the spiral the spiral arms would be 3.14" apart. What would B be if the spiral arms were 0.65625" apart? 2(pi)B=0.65625, so B=0.65625/(2(pi))=0.10445". The equation of the spiral is r=5+0.10445ø. To calculate the length of the spiral we have two possible ways: an approximate value based on the similarity between concentric circles and a spiral; or an accurate value obtainable through calculus. The approximate way is to add together the circumferences of the concentric circles: L=2(pi)(5+(5+0.65625)+...+(5+0.65625N)) where L=spiral length and N is the number of turns. L=2(pi)(5N+0.65625S) where S=0+1+2+3+...+(N-1)=N(N-1)/2. This formula arises from the fact that the first and last terms (0, N-1) the second and penultimate terms (1, N-2) and so on add up to N-1. So, for example, if N were 10 we would have (0+9)+(1+8)+(2+7)+(3+6)+(4+5)=5*9=45=10*9/2. If N were 5 we would have 0+1+2+3+4=10=(0+4)+(1+3)+2=5*4/2. L=12*100 inches. L=1200=2(pi)(5N+0.65625N(N-1)/2)=(pi)N(10+0.65625(N-1))=(pi)N(9.34375+0.65625N). If the external radius is r1 and the internal radius is r then the thickness of the spiral is r1-r and since 0.65625 is the gap between the spiral arms N=(r1-r)/0.65625. N is an integer, but, since it is unlikely that this equation would actually produce an integer we would settle for the nearest integer. If we solve this equation for N, we can deduce the external radius and diameter of the spiral: N(9.34375+0.65625N)=1200/(pi)=381.97; 0.65625N^2+9.34375N-381.97=0 and N=(-9.34375+sqrt(1089.98))/1.3125=18 (nearest integer). This means that there are 18 turns of the spiral to make the total length about 100 feet. If X is the final external diameter of the coiled pipe and the internal radius is 5" (the minimum allowable) then X/2 is the external radius, so N=((X/2)-5)/0.65625. We found N=18 so we can find X: X=2*(0.65625*18+5)=33.625in. Solution using calculus Using calculus, we can work out the relationship between the length of the spiral and other parameters. We start with any polar equation r(ø) and a picture: draw a line representing a general value of r. At a small angle dø to this line we draw another line a little bit longer, length r+dr. Now we join the ends together to make a narrow-angled triangle AOB where angle AOB=dø and AB=ds, the small section of the curve. In the triangle AO is length r and BO is length r+dr. If we mark the point C along BO so that CO is length r, the same as AO, we have an isosceles triangle COA. Because the apex angle is small, CA=rdø, the length of the arc of the sector. In triangle ABC, CB=dr, AB=ds and CA=rdø. By Pythagoras, AB^2=CB^2+CA^2, that is, ds^2=dr^2+r^2dø^2, because angle BCA is a right angle as dø tends to zero. The length of the curve is the result of adding the tiny ds values together between limits of r or ø. We can write ds=sqrt(dr^2+r^2dø^2). If we divide both sides by dr, we get ds/dr=sqrt(1+(rdø/dr)^2) so s=integral(sqrt(1+(rdø/dr)^2)dr, where s is the length of the curve. The integral is definite if we define the limits of r. For our spiral we have r=A+Bø, making ø=(r-A)/B and B=p/(2(pi)), where p is the diameter of the pipe=0.65625", so we can substitute for ø in the integral and the limits for r are A to X/2, where A is the inner radius (A=5") and X/2 is the outer radius. dø/dr=2(pi)/p, a constant=9.57 approx. s=integral(sqrt(1+(2(pi)r/p)^2)dr) between limits r=A to X/2. After the integral is calculated, we solve for X putting s=1200". The expression (2(pi)r/p)^2 is large compared to 1, so s=integral((2(pi)r/p)dr) approximately and s=[(pi)r^2/p] (r=A to X/2); therefore, since we know s=1200, we can write ((pi)/p)(X^2/4-A^2)=1200. Therefore X=2sqrt(1200p/(pi))+A^2)=33.21". Compare this answer with the one we got before and we can see they are close. [We could get a formal solution to the integral, using hyperbolic trigonometric or other logarithmic functions, but such a solution would make it very difficult or tedious to solve for X, since X would appear in logarithmic expressions and in other expressions making it difficult or impossible to isolate X. For example, the next term in the expansion of the integral would be (p/(4(pi))ln(X/2A), having a value of about 0.06. It is anticipated, therefore, that an approximation would be sufficient in this problem with the given figures.] We can feel justified in using the formula for finding the length of pipe, L, when X=6'=72": L=((pi)/p)(1296-25)=6084.52"=507' approximately. This length of pipe would hold 507/100*0.96 gallons=4.87 gallons.      
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how to find perimeter of a joint figure

To find the perimeter of a joint figure first find out what simpler figures the whole figures is made of. For example, two right-angled triangles and a rectangle all of equal height can be joined together to make a trapezium, or trapezoid. You then need to identify the common edges, because these will not form part of the perimeter, and you need to identify the exposed edges, because these will form part of the perimeter. So going back to the trapezium, the heights of the triangles and two sides of the rectangle are common, so these will be excluded, but the hypotenuses of the triangles and the other two sides are exposed, so they form part of the perimeter.  You need to know the lengths of the exposed edges. Although the common height is not part of the perimeter, you may need it to work out the lengths of the exposed sides or edges. For the triangles you need the lengths of the hypotenuses, so, using whatever information you have about the triangles, find the lengths of the hypotenuses. You may need Pythogars' theorem or trigonometry to do this, depending on whether you are given the lengths of sides or angle measurements. The rectangle length or width will be exposed, and so you need to work out their length from the info you have. The perimeter will be the sum of the two hypotenuses and bases of the triangles, plus twice the length (or width) of the rectangle. Another example is a triangle attached to a semicircle, so that one side of the triangle is a diameter of the semicircle. The length of the circumference of a circle is 2(pi)r, where r is the radius, so (pi)r is the exposed part of the semicircle. You need the lengths of the exposed sides of the triangle. 2r is the length of one side, and you need more info about the triangle to work out the length of the other two sides. Combine these to get the perimeter of the whole figure. These are just examples. You need to examine closely how your joint figure has been made. You might like to work out the perimeter of a regular 5-pointed star built on a regular pentagon. Here you have a pentagon with five isosceles triangles sitting on its sides, but none of the sides of the pentagon are exposed and two sides of each triangle, all with the same length are exposed, so the perimeter is ten times the length of just one side of a triangle. You need info about the pentagon to work out the length of this side.
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A math word problem using the Quadratic Equation formula (URGENT)

A. The Golden Ratio I have a line divided into two unequal sections so that the ratio of the shorter length to the longer length is the same as the ratio of the longer length to the whole length of the line. What is the ratio? Solution Call the lengths of the two sections A and B. A/B=B/(A+B). Cross-multiply: A^2+AB=B^2, so A^2+AB-B^2=0 and, using the quadratic formula, A=(-B+sqrt(B^2+4B^2)/2=(-B+Bsqrt(5))/2=B(-1+sqrt(5))/2. Therefore the fraction A/B=(sqrt(5)-1)/2=0.618 or -1.618 approx. The positive root applies because A and B are considered positive lengths. An alternative solution is to let r=A/B, which is the ratio, then r=1/(r+1); r^2+r-1=0 and r=(sqrt(5)-1)/2.  This Golden Ratio keeps popping up in different contexts and seems to be aesthetically pleasing. B. The A sizes A rectangular piece of paper is folded in half so that the ratio of length to width remains the same after folding as before folding. What is the ratio? Solution If A and B are the length and width, the area is AB, and after folding in half the area becomes 1/2(AB). In the process of folding, only one side (choose the length A) is halved so the new dimensions of the rectangle are A/2 and B, but the ratio of the sides is still A/B, and the length and width are B and A/2 (the length has become the width and the width has become the length). So A/B=length/width=B/(A/2)=2B/A. If A/B=2B/A then A^2=2B^2. The ratio of the sides is A/B which we'll call r. So r^2=2 and r=sqrt(2)=1.414 approx. So the length is 1.414 times the width. The paper we use today is largely based on this ratio. A4 is the standard size used for printers. A3 is larger and tends to be used for drawings and posters. A5 is smaller and tends to be used for booklets and leaflets. All use the square root of 2 as the ratio of the sides. C. Identity Prove that the difference between the cubes of two consecutive integers is one more than three times their product. Solution The difference between the cubes of consecutive integers (x+1)^3-x^3 where x is an integer. The product of consecutive integers is x(x+1). The difference between the cubes is x^3+3x^2+3x+1-x^3=3x^2+3x+1=3x(x+1)+1 which is three times the product of the integers plus 1. These questions use in different ways a quadratic expression. Although this reply is later than you wished for, what goes round comes round, and you may have the opportunity of using the ideas.
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Plz it needs help plz

Question: The mean length of 6 rods is 44.2cm The mean length of 5 of them is 46cm.How long is the sixth one? Since the mean length of 6 rods is 44.2cm, then their total length is 6*44.2 = 265.2 cm. And since the mean length of 5 of them is 46cm, then the total length of these 5 rods is 5*46 = 230 cm. So the difference between total lengths, 265.2 - 230 = 35.2 cm, is the length of the 6th rod. Answer: length of the 6th rod is 35.2 cm
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An alligators tail length T

An alligator’s tail length, T varies directly as its body length, B and an alligator with a body length of 4 ft has a tail of 2.5 ft. so what is the tail length of an alligator whose body length is 4.8 ft? 4:2.5 = 4.8:T 1.6 = 4.8T 3 = T
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the perimeter of a rectangle is 66cm and it's width is half its length. what are the length and width of the rectangle

First let's state the formula for perimeter of a rectangle: Perimeter = 2(length) + 2(width) OK, for this rectangle let's set the length equal to x length = x Now we are told that the width is one half it length.  So if the length is x then width = 1/2*(x) or x/2 OK, the problem also tells us that the perimeter is 66 cm Now we have everything we need to use the perimeter formula and solve for x Perimeter = 2(length) + 2(width) 66 = 2(x) + 2(x/2) 66 = 2x + x 66 = 3x 22 = x Answer: the length is 22 cm Now use this value for length to find the width width = x/2 width = 22/2 width = 11 Answer: the width is 11 cm  Check both answers by using the perimeter formula. Perimeter = 2(length)+2(width)  66 = 2(22) + 2(11) 66 = 44 + 22 66 = 66 
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you can use rods of integer sizes to build trains that all share a common length. a train of length 5 is a row of rods whose combined length is many trains of length 5 are there?

The integer-sized rods would consist of 1, 2, 3, 4 and 5. To build a train of length 5 you could use 1 rod; two rods of lengths 1 and 4, 2 and 3; three rods of lengths (1, 1 and 3) and (1, 2 and 2); four rods of 1, 1, 1 and 2; and five rods of lengths 1, 1, 1, 1 and 1. If you count, for example, 1 and 4 as being different from 4 and 1, then there would be 4 ways of using two rods; six ways of using three rods; four ways of using four rods; and, of course, one way of using five rods or one rod. This pattern 1 4 6 4 1 is a row of Pascal's triangle and is related to n, where n is the length of the train. When we add these numbers together we get 16, which is 2^4. For a train of length n, my guess is that the total number of ways of setting up the rods is 2^(n-1).
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Find the volume and surface area ?

1) Volume = length * breadth * height = 8 cm * 2.5 cm * 2 cm = 40 cm^3 Surface area = 2(length * breadth) + 2(breadth * height) + 2(length * height) = 2(8 cm * 2.5 cm) + 2(2.5 cm * 2 cm) + 2(8 cm * 2 cm) = 2(20 cm^2) + 2(5 cm^2) + 2(16 cm^2) = 40 cm^2 + 10 cm^2 + 32 cm^2 = 82 cm^2 2) Weird length here. Length is usually more than breadth. I will just follow with 1.5cm, and I hope it is not 1.5m. Volume = length * breadth * height = 1.5 cm * 90 cm * 70 cm = 9450 cm^2 Surface area = 2(length * breadth) + 2(breadth * height) + 2(length * height) = 2(1.5 cm * 90 cm) + 2(90 cm * 70 cm) + 2(1.5 cm * 70 cm) = 2(135 cm^2) + 2(6300 cm^2) + 2(105 cm^2) = 270 cm^2 + 12600 cm^2 + 210 cm^2 = 13080 cm^2
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Find the length and volume of the wire

The copper wire will form a tight spiral around the cylinder, but we can approximate to this by treating the wire as a number of segments. Each segment has a length equal to the circumference of the cylinder, which is 49π cm (about 3 times its diameter). The segments are placed side by side so we can work out how many there are by dividing the diameter of the wire into the length of the cylinder: 18/0.6=30. So we now know that we need 30 segments of wire, where each segment is 49π cm long. The volume of the segment of wire is length times cross-sectional area, π(0.3)^2=0.09π sq cm, so the volume of a segment is 49π*0.09π cc=43.5250cc. There are 30 of these side by side so the total volume is 30*43.5250=1305.75cc. The length of wire is 30*49π=4618.14cm. This is a fair approximation, because each segment will actually be slightly longer so as to join up with the next segment to form a continuous length, but because the cylinder's dimensions are very much greater than the wire's as a cylinder, the difference won't be significant. So 4618cm and 1306cc should be accurate enough for the length and volume of wire.
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